I've been thinking more about your tidal theory, JohnD, and I think it's largely correct as far as the solar tides are concerned but requires some modification to explain lunar tides. The point I want to make in this post is that the rotation of the Earth about its axis is largely irrelevant to the question of solar tides, but is vitally important in explaining lunar tides.

Solar Tides: Let's ignore the Moon for the moment and start with solar tides. In his book the BA says that he relied heavily on Mikolaj Sawicki's Myths About Gravity and Tides. Sawicki's explanation of solar tides, you'll be happy to know, is exactly the one you gave.

The Earth's centre of mass (CoM) is in free-fall about the barycentre of the Earth-Sun system (which is located near the centre of the Sun), orbiting at 108,000 kph, just the right speed to stay in orbit.

The point on the Earth's surface closest to the Sun is also travelling about the Earth-Sun barycentre at this speed, but being closer it should be travelling faster (Kepler's third law) to be in orbit; so it has a tendency to slip into the Sun's gravity well, and this accounts for the tidal bulge on that side of the Earth.

Meanwhile, the point on the Earth's surface farthest from the Sun is travelling at 108,000 kph, but being farther from the Sun it should be travelling more slowly to be in free-fall; so it has a tendency to climb up out of the Sun's gravity well, and this accounts for the solar tidal bulge on the far side of the Earth.

In all of this, the rotation of the Earth about its own axis is not very relevant. A point on the equator is travelling around the centre of the Earth at a speed of about 2,000 kph, which is only about one fiftieth of the Earth's orbital velocity. This will reduce the orbital speed of the point closest to the Sun to about 106,000 kph, and increase the orbital speed of the point farthest from the Sun to about 110,000 kph. So the rotation of the Earth will enhance the tidal effect, but only slightly.

Lunar Tides: Now let's look at the Moon. After explaining solar tides, Sawicki says:
Unfortunately he does not go into any greater detail than this. I contend that the lunar tidal situation differs from the solar one in two crucial ways:

[1] With solar tides, the Earth's two tidal bulges are always on the same side of the Earth-Sun barycentre. With lunar tides, they're always on opposite sides of the Earth-Moon barycentre (which is located 1,600 km beneath the Earth's surface).

[2] With solar tides, the Earth's speed of axial rotation (2,000 kph) is 50 times smaller than the Earth's orbital velocity (108,000 kph). With lunar tides, it's 50 times greater.

The Earth's CoM orbits the Earth-Moon barycentre in 27.32 days. So its orbital speed about the barycentre is a pedestrian 44 kph (do the math!). This means that the point on the Earth's surface closest to the Moon is actually orbiting the Earth-Moon barycentre at about 2,044 kph. To be in free-fall it should be orbiting at 75 kph (do the math!). Because of the Earth's axial rotation, it's actually travelling at 2,040 kph, so it has a tendency to move away from the barycentre. This accounts for the tidal bulge on the near side of the Earth.

A similar situation obtains on the far side. For free-fall, that point should be travelling around the barycentre at 29 kph (again, I leave you to check these figures). But it's actually travelling round it at 2,044 kph, so it too has a tendency to climb into a higher orbit. This accounts for the tidal bulge on the far side of the Earth.

There are two things about this explanation which worry me, though:

[1] Neither Sawicki nor the BA mention any of this in their explanations. As they're the experts and I'm the plodder, I suspect that I have made some fundamental error in all of this.

[2] As kilopi pointed out, the two tidal bulges are the same size. Can my theory account for that? I don't know. Guess I'll have to crunch some numbers, but I'll leave that for another post.

Phew!

__________________
- Learn a lot teaching others.